For the function $\frac{1}{x}$ on the real line, one can use a modified principal value integral to consider it as a distribution p.f.$(\frac{1}{x}),$ and one can do a similar construction to make $\frac{1}{x^m}$ into a distribution for $m>1.$ In the complex plane, the function $\frac{1}{z^m}$ is locally integrable for $m=1,$ but for larger $m$ some construction analogous to the one dimensional would have to be done to make it into a distribution.

More generally, given a meromorphic function on the plane (or torus), one should be able to consider it as a distribution by integrating against it and subtracting off some delta distributions or derivatives of delta distributions. Is this process explained in detail anywhere? Has anyone computed the Fourier series of such distributions, say for the Weierstrass $\mathfrak{p}$ function on the torus?

Concerning your first question: By taking $m-1$-th distributional derivatives of the locally integrable function $1/z$ one can extend $z^{-m}$ from the punctured complex plane to a distribution on the full complex plane.
–
Sönke HansenApr 18 '13 at 21:32

2 Answers
2

It is folklore that any meromorphic function on the real line can be regarded as a distribution in a natural way. One defines $x^{-n}$ to be the $n$-th derivative of the locally integrable function $\log |x|$, with appropriate coefficient. In order to treat the general case, one uses the principle of recollement des morceaux---if the real line is covered by a family of open subsets and one has a family of distributions, one on each set, which satisfies the obvious compatibility condition, then one can combine them to a single distribution. This is a theorem in the Schwartz book---the proof uses partitions of unity. I heard this argument in a course given regularly by the portuguese mathematician J. Sebastião e Silva at the University of Lisbon (I attended it in 1969).
The same globalisation argument, combined with the comment of Sönke Hansen above, gives the result for the complex plane. I have no information on the second part of the query.

I forgot one more ingredient---you can always multiply a distribution with a smooth function. So you consider an open cover which is such that each element contains only one pole. The restriction of the meromorphic function to such a set is an analytic function times one of the form $(x-x_0)^{-n}$ and so a distribution. Now globalise.
–
jbcApr 18 '13 at 22:27

For a doubly periodic meromorphic function (so a function on the torus) does this gluing procedure ensure that the resulting distribution will also be doubly periodic? It seems like it should, since in any neighborhood not containing a pole it is given by integration against a doubly periodic function.
–
mkreiselApr 19 '13 at 14:25

It's interesting to think distributiions as the boundary of analytic functions, which I believe is originated from Saito. You may find the following result helpful

Let I be an open interval on $\mathbb{R}$, and
$$
Z={z\in \mathbb{C};\Re z\in I,0<\Im z <\gamma}
$$
is a one sided complex neighborhood. If $f$ is analytic such that there exsists a non-negtive integer $N$
$$
|f(z)|\leq C|\Im z|^{-N},\quad z\in Z
$$
Then $f(x+i0)$ exsists as a distribution and is of order $N+1$. And the condition can not be relax very much, in fact, if we assume that $f(x+i0)\in \mathcal{D'}^n$(distribution with order $n$),then we have
$$
|f(z)|\leq C|\Im z|^{-n-1}
$$